Question: $\overline{AC}$ is $3$ units long $\overline{BC}$ is $10$ units long $\overline{AB}$ is $\sqrt{109}$ units long What is $\cos(\angle ABC)$ ? $A$ $C$ $B$ $3$ $10$ $\sqrt{109}$
Solution: SOH CAH TOA os = djacent over ypotenuse adjacent $= \overline{BC} = 10$ hypotenuse $= \overline{AB} = \sqrt{109}$ $\cos(\angle ABC )=\frac{10}{\sqrt{109}}$ $=\dfrac{10\sqrt{109} }{109}$